Electron configuration of ionized Copper

Question

I am trying to think about the electron configuration of copper by adding electrons one by one, starting from a bare copper nucleus. My understanding is that the violation of the Aufbau principle occurs right as we add the last electron, as shown below:

0. Suppose we have already added $29-2=27$ electrons to the nucleus.

1. Before adding the last two electrons, we have $\text{Cu}^{2+}=[\text{Ar}]3\text{d}^7 4\text{s}^2$

2. Before adding the last electron, we have $\text{Cu}^{1+}=[\text{Ar}]3\text{d}^8 4\text{s}^2$

3. After adding the last electron, we have $\text{Cu} = [\text{Ar}]3\text{d}^{10} 4\text{s}^1$

Why then, is the electron configuration of ionized copper, $\text{Cu}^{1+}$, in reality equal to $[\text{Ar}]3\text{d}^{10}$? Is it not energetically favorable for two of the $3\text{d}$ electrons to go to $4\text{s}$ after ionization?

Answer 1

This can be explained via the (slightly empirical) concepts of shielding.

In the one-electron ion Cu$^{28+}$, orbitals with the same $n$ have the same energy (neglecting some small effects). Thus, the $3d$ orbital will lie below $4s$.

As you add electrons, the energy of $3d$ will increase relative to $4s$. This is because the latter has some electron density close to the nucleus, so it still feels some of the original nuclear charge, whereas $3d$ has its electron density quite far away - it is thus very well shielded by the core electrons.

This argument is only a qualitative explanation. However, by actually solving the Schrödinger equation (or rather, finding an approximation solution like the Hartree-Fock method), energies of the orbitals may be derived and it is possible to show where the crossover between occupation of $4s$ and $3d$ occurs.

This matter is further complication in copper by the exchange interaction, which stabilises $3d^{10}$ relative to $3d^9 4s^1$.